If $A$ is a commutative $C^*$ $*$-algebra, then the kernel of a nontrivial character is a maximal ideal.
If $A$ is an arbitrary $C^*$-algebra, what kind of subspace is the kernel of a pure state?
The motivation is the following: in the commutative case, we have the $C^*$-algebra version of the Stone-Weierstrass theorem, in which the concept of "character" (or, equivalently: maximal ideal) is central; if the algebra is no longer commutative, the space of characters (the spectrum of the algebra) is replaced by the pure state space. I would therefore like to understand what the correspondence between characters and maximal ideals changes into when the underlying algebra is no longer commutative.
With the following two posts:
$\tau$ is a pure state and $\tau(a)=0$. Then $a=b+c$ for some $b,c$ with $\tau (b^*b)=\tau(cc^*)=0$.
https://mathoverflow.net/questions/223588
I believe you can conclude a state is pure iff its kernel is equal to $I+I*$ where $I$ is a left ideal. The first direction is obvious. To show the converse, according to the second post, we have that $I$ is the kernel of a state and hence, by the first post, we know that state is pure.